Physics - Unit & Dimension Question with Solution | TestHub

Dimensional formula of surface tension, $\left[S\right]=\left[{\mathrm{M}}^1{\mathrm{L}}^0{\mathrm{T}}^{-2}\right]$
Dimensional formula of energy, $\left[E\right]=\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]$
Dimensional formula of velocity, $\left[V\right]=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-1}\right]$

By applying dimensional analysis,
$\mathrm{S}={\mathrm{E}}^{\mathrm{a}}{\mathrm{V}}^{\mathrm{b}}{\mathrm{T}}^{\mathrm{c}}$
$\left[{\mathrm{M}}^1{\mathrm{L}}^0{\mathrm{T}}^{-2}\right]={\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]}^{\mathrm{a}}{\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-1}\right]}^{\mathrm{b}}{\left[\mathrm{T}\right]}^{\mathrm{c}}$
$\left[{\mathrm{M}}^1{\mathrm{L}}^0{\mathrm{T}}^{-2}\right]={\mathrm{M}}^{\mathrm{a}}{\mathrm{L}}^{2\mathrm{a}+\mathrm{b}}{\mathrm{T}}^{-2\mathrm{a}-\mathrm{b}+\mathrm{c}}$

On comparing,
$\mathrm{a}=1$
$2\mathrm{a}+\mathrm{b}=0\Rightarrow \mathrm{b}=\mathrm{ }-2$
$-2\mathrm{a}-\mathrm{b}+\mathrm{c}=-2\Rightarrow \mathrm{c}=\mathrm{ }-2$
$\therefore \mathrm{S}={\mathrm{E}}^1{\mathrm{V}}^{-2}{\mathrm{T}}^{-2}$